Colligative
properties calculations are used for this type of problem. Calculations are as
follows:<span>
</span>
<span>ΔT(freezing point)
= (Kf)m
ΔT(freezing point)
= 1.86 °C kg / mol (0.705)
ΔT(freezing point) = 1.3113 °C
</span>
<span>
</span>
<span>Hope this answers the question. Have a nice day.</span>
Answer:
See explanation below
Explanation:
In order to calculate this, we need to use the following expression to get the concentration of the base:
MaVa = MbVb (1)
We already know the volume of NaOH used which is 13.4473 mL. We do not have the concentration of KHP, but we can use the moles. We have the mass of KHP which is 0.5053 g and the molecular formula. Let's calculate the molecular mass of KHP:
Atomic weights of the elements to be used:
K = 39.0983 g/mol; H = 1.0078 g/mol; C = 12.0107 g/mol; O = 15.999 g/mol
MM KHP = (1.0078*5) + (39.0983) + (8*12.0107) + (4*15.999) = 204.2189 g/mol
Now, let's calculate the mole of KHP:
moles = 0.5053 / 204.2189 = 0.00247 moles
With the moles, we also know that:
n = M*V (2)
Replacing in (1):
n = MbVb
Now, solving for Mb:
Mb = n/Vb (3)
Finally, replacing the data:
Mb = 0.00247 / (13.4473/1000)
Mb = 0.184 M
This would be the concentration of NaOH
I believe the answer is gills
Fish's use gills to breathe, so tadpoles should use gills too
hope this helps
(got the info from a documentary i watched a couple years ago)
2AgCl (s) + Ba(NO3)2 (aq)
